Question

Solve (3y + 2x + 4) * d * x - (4x + 6y + 5) * d * y = 0

Answer 1

$y'=\frac{3y+2x+4}{6y+4x+5}\\ y'=\frac{3y+2x+4}{2(3y+2x)+5}$

substitute $3y+2x=z(x) \to y=\frac{z-2x}{3} \to y'=\frac{z'-2}{3}$

then

$\frac{z'-2}{3}=\frac{z+4}{2z+5}\\ z'-2=\frac{3z+12}{2z+5}\\ z'=\frac{7z+22}{2z+5}\\ \frac{2z+5}{7z+22}dz=dx\\ \frac{1}{7}\int(\frac{2(7z+22)-\frac{22}{7} +35}{7z+22}dz=\int dx\\ \frac{1}{7}\int(2+\frac{223}{7}\cdot \frac{1}{7z+22})dz=\int dx\\ \frac{1}{7}(2z+\frac{223}{49} \ln|7z+22|)=x+c\\$

Answer

$\frac{1}{7}(2(3y+2x)+\frac{223}{49} \ln|7(3y+2x)+22|)=x+c$